midterm 2 revision
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Midterm 2 RevisionMidterm 2 Revision
Prof. Sin-Min Lee
Department of Computer Science
San Jose State University
©Silberschatz, Korth and Sudarshan3.2Database System Concepts
1. Know the difference between a database and a DBMS
Functions/advantages and disadvantages of a DBMS
2.Understand the meaning of all of the E-R symbols .
3.Know the basis of the mathematical relation and the properties of a relation. Understand and recognize symbols for Selection, projection, Cartesian product, union and set difference. Understand the difference between an inner join and an outerjoin
4.Know the characteristics of superkey, candidate key, primary key, and foreign key.
5.Know the rules of relational integrity and referential integrity.
6. Be able to recognize and read relational algebra statements with the primary operators.
7.Be able to recognized simple relational calculus statements (like the ones used in class) and understand the difference between the algebra and calculus.
Materials cover in Exam.
©Silberschatz, Korth and Sudarshan3.3Database System Concepts
Multiple Choice Problems
©Silberschatz, Korth and Sudarshan3.4Database System Concepts
©Silberschatz, Korth and Sudarshan3.5Database System Concepts
©Silberschatz, Korth and Sudarshan3.6Database System Concepts
E-R Diagram for the Banking EnterpriseE-R Diagram for the Banking Enterprise
©Silberschatz, Korth and Sudarshan3.7Database System Concepts
Determining Keys from E-R SetsDetermining Keys from E-R Sets
Strong entity set. The primary key of the entity set becomes the primary key of the relation.
Weak entity set. The primary key of the relation consists of the union of the primary key of the strong entity set and the discriminator of the weak entity set.
Relationship set. The union of the primary keys of the related entity sets becomes a super key of the relation. For binary many-to-one relationship sets, the primary key of the
“many” entity set becomes the relation’s primary key.
For one-to-one relationship sets, the relation’s primary key can be that of either entity set.
For many-to-many relationship sets, the union of the primary keys becomes the relation’s primary key
©Silberschatz, Korth and Sudarshan3.8Database System Concepts
Chapter 3: Relational ModelChapter 3: Relational Model
Structure of Relational Databases
Relational Algebra
Tuple Relational Calculus
Domain Relational Calculus
Extended Relational-Algebra-Operations
©Silberschatz, Korth and Sudarshan3.9Database System Concepts
Basic StructureBasic Structure
Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn
Thus a relation is a set of n-tuples (a1, a2, …, an) where ai Di
Example: if
customer-name = {Jones, Smith, Curry, Lindsay}customer-street = {Main, North, Park}customer-city = {Harrison, Rye, Pittsfield}
Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield)} is a relation over customer-name x customer-street x customer-city
©Silberschatz, Korth and Sudarshan3.10Database System Concepts
Attribute TypesAttribute Types
Each attribute of a relation has a name
The set of allowed values for each attribute is called the domain of the attribute
Attribute values are (normally) required to be atomic, that is, indivisible E.g. multivalued attribute values are not atomic
E.g. composite attribute values are not atomic
The special value null is a member of every domain
The null value causes complications in the definition of many operations we shall ignore the effect of null values in our main presentation
and consider their effect later
©Silberschatz, Korth and Sudarshan3.11Database System Concepts
Relation SchemaRelation Schema
A1, A2, …, An are attributes
R = (A1, A2, …, An ) is a relation schema
E.g. Customer-schema = (customer-name, customer-street, customer-city)
r(R) is a relation on the relation schema R
E.g. customer (Customer-schema)
©Silberschatz, Korth and Sudarshan3.12Database System Concepts
Relation InstanceRelation Instance The current values (relation instance) of a relation are
specified by a table
An element t of r is a tuple, represented by a row in a table
JonesSmithCurry
Lindsay
customer-name
MainNorthNorthPark
customer-street
HarrisonRyeRye
Pittsfield
customer-city
customer
attributes
tuples
©Silberschatz, Korth and Sudarshan3.13Database System Concepts
Relations are UnorderedRelations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
E.g. account relation with unordered tuples
©Silberschatz, Korth and Sudarshan3.14Database System Concepts
KeysKeys
Let K R
K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) by “possible r” we mean a relation r that could exist in the enterprise we are modeling.Example: {customer-name, customer-street} and {customer-name} are both superkeys of Customer, if no two customers can possibly have the same name.
K is a candidate key if K is minimalExample: {customer-name} is a candidate key for Customer, since it is a superkey {assuming no two customers can possibly have the same name), and no subset of it is a superkey.
©Silberschatz, Korth and Sudarshan3.15Database System Concepts
Query LanguagesQuery Languages
Language in which user requests information from the database.
Categories of languages procedural
non-procedural
“Pure” languages: Relational Algebra
Tuple Relational Calculus
Domain Relational Calculus
Pure languages form underlying basis of query languages that people use.
©Silberschatz, Korth and Sudarshan3.16Database System Concepts
Relational AlgebraRelational Algebra
Procedural language
Six basic operators select
project
union
set difference
Cartesian product
rename
The operators take two or more relations as inputs and give a new relation as a result.
©Silberschatz, Korth and Sudarshan3.17Database System Concepts
Select Operation – ExampleSelect Operation – Example
• Relation r A B C D
15
1223
773
10
A=B ^ D > 5 (r)A B C D
123
710
©Silberschatz, Korth and Sudarshan3.18Database System Concepts
Select OperationSelect Operation
Notation: p(r) p is called the selection predicate Defined as:
p(r) = {t | t r and p(t)}
Where p is a formula in propositional calculus consisting of terms connected by : (and), (or), (not)Each term is one of:
<attribute> op <attribute> or <constant>
where op is one of: =, , >, . <. Example of selection:
branch-name=“Perryridge”(account)
©Silberschatz, Korth and Sudarshan3.19Database System Concepts
Project Operation – ExampleProject Operation – Example
Relation r: A B C
10203040
1112
A C
1112
=
A C
112
A,C (r)
©Silberschatz, Korth and Sudarshan3.20Database System Concepts
Project OperationProject Operation
Notation:
A1, A2, …, Ak (r)
where A1, A2 are attribute names and r is a relation name.
The result is defined as the relation of k columns obtained by erasing the columns that are not listed
Duplicate rows removed from result, since relations are sets
E.g. To eliminate the branch-name attribute of account account-number, balance (account)
©Silberschatz, Korth and Sudarshan3.21Database System Concepts
Union Operation – ExampleUnion Operation – Example
Relations r, s:
r s:
A B
121
A B
23
rs
A B
1213
©Silberschatz, Korth and Sudarshan3.22Database System Concepts
Union OperationUnion Operation
Notation: r s
Defined as:
r s = {t | t r or t s}
For r s to be valid.
1. r, s must have the same arity (same number of attributes)
2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s)
E.g. to find all customers with either an account or a loan customer-name (depositor) customer-name (borrower)
©Silberschatz, Korth and Sudarshan3.23Database System Concepts
Set Difference Operation – ExampleSet Difference Operation – Example
Relations r, s:
r – s:
A B
121
A B
23
rs
A B
11
©Silberschatz, Korth and Sudarshan3.24Database System Concepts
Set Difference OperationSet Difference Operation
Notation r – s
Defined as:
r – s = {t | t r and t s}
Set differences must be taken between compatible relations. r and s must have the same arity
attribute domains of r and s must be compatible
©Silberschatz, Korth and Sudarshan3.25Database System Concepts
Cartesian-Product Operation-ExampleCartesian-Product Operation-Example
Relations r, s:
r x s:
A B
12
A B
11112222
C D
1019201010102010
E
aabbaabb
C D
10102010
E
aabbr
s
©Silberschatz, Korth and Sudarshan3.26Database System Concepts
Cartesian-Product OperationCartesian-Product Operation
Notation r x s
Defined as:
r x s = {t q | t r and q s}
Assume that attributes of r(R) and s(S) are disjoint. (That is, R S = ).
If attributes of r(R) and s(S) are not disjoint, then renaming must be used.
©Silberschatz, Korth and Sudarshan3.27Database System Concepts
Composition of OperationsComposition of Operations
Can build expressions using multiple operations
Example: A=C(r x s)
r x s
A=C(r x s)
A B
11112222
C D
1019201010102010
E
aabbaabb
A B C D E
122
102020
aab
©Silberschatz, Korth and Sudarshan3.28Database System Concepts
Rename OperationRename Operation
Allows us to name, and therefore to refer to, the results of relational-algebra expressions.
Allows us to refer to a relation by more than one name.
Example:
x (E)
returns the expression E under the name X
If a relational-algebra expression E has arity n, then
x (A1, A2, …, An) (E)
returns the result of expression E under the name X, and with the
attributes renamed to A1, A2, …., An.
©Silberschatz, Korth and Sudarshan3.29Database System Concepts
Banking ExampleBanking Example
branch (branch-name, branch-city, assets)
customer (customer-name, customer-street, customer-only)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)
©Silberschatz, Korth and Sudarshan3.30Database System Concepts
Example QueriesExample Queries
Find all loans of over $1200
amount > 1200 (loan)
Find the loan number for each loan of an amount greater than $1200
loan-number (amount > 1200 (loan))
©Silberschatz, Korth and Sudarshan3.31Database System Concepts
Example QueriesExample Queries
Find the names of all customers who have a loan, an account, or both, from the bank
customer-name (borrower) customer-name (depositor)
Find the names of all customers who have a loan and an account at bank.
customer-name (borrower) customer-name (depositor)
©Silberschatz, Korth and Sudarshan3.32Database System Concepts
Example QueriesExample Queries
Find the names of all customers who have a loan at the Perryridge branch.
customer-name (branch-name=“Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan)))
Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank.
customer-name (branch-name = “Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan)))
– customer-name(depositor)
©Silberschatz, Korth and Sudarshan3.33Database System Concepts
Example QueriesExample Queries
Find the names of all customers who have a loan at the Perryridge branch.
Query 1
customer-name(branch-name = “Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan)))
Query 2
customer-name(loan.loan-number = borrower.loan-number(
(branch-name = “Perryridge”(loan)) x
borrower) )
©Silberschatz, Korth and Sudarshan3.34Database System Concepts
Example QueriesExample Queries
Find the largest account balance
Rename account relation as d
The query is:
balance(account) - account.balance
(account.balance < d.balance (account x d (account)))
©Silberschatz, Korth and Sudarshan3.35Database System Concepts
Additional OperationsAdditional Operations
We define additional operations that do not add any power to the
relational algebra, but that simplify common queries.
Set intersection
Natural join
Division
Assignment
©Silberschatz, Korth and Sudarshan3.36Database System Concepts
Set-Intersection OperationSet-Intersection Operation
Notation: r s
Defined as:
r s ={ t | t r and t s }
Assume: r, s have the same arity
attributes of r and s are compatible
Note: r s = r - (r - s)
©Silberschatz, Korth and Sudarshan3.37Database System Concepts
Set-Intersection Operation - ExampleSet-Intersection Operation - Example
Relation r, s:
r s
A B
121
A B
23
r s
A B
2
©Silberschatz, Korth and Sudarshan3.38Database System Concepts
Natural-Join OperationNatural-Join Operation
Notation: r s Let r and s be relations on schemas R and S respectively.The result is a
relation on schema R S which is obtained by considering each pair of tuples tr from r and ts from s.
If tr and ts have the same value on each of the attributes in R S, a tuple t is added to the result, where
t has the same value as tr on r
t has the same value as ts on s
Example:
R = (A, B, C, D)
S = (E, B, D) Result schema = (A, B, C, D, E) r s is defined as:
r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r x s))
©Silberschatz, Korth and Sudarshan3.39Database System Concepts
Natural Join Operation – ExampleNatural Join Operation – Example
Relations r, s:
A B
12412
C D
aabab
B
13123
D
aaabb
E
r
A B
11112
C D
aaaab
E
s
r s
©Silberschatz, Korth and Sudarshan3.40Database System Concepts
Division OperationDivision Operation
Suited to queries that include the phrase “for all”. Let r and s be relations on schemas R and S respectively
where R = (A1, …, Am, B1, …, Bn)
S = (B1, …, Bn)
The result of r s is a relation on schema
R – S = (A1, …, Am)
r s = { t | t R-S(r) u s ( tu r ) }
r s
©Silberschatz, Korth and Sudarshan3.41Database System Concepts
Division Operation – ExampleDivision Operation – Example
Relations r, s:
r s: A
B
12
A B
12311134612
r
s
©Silberschatz, Korth and Sudarshan3.42Database System Concepts
Another Division ExampleAnother Division Example
A B
aaaaaaaa
C D
aabababb
E
11113111
Relations r, s:
r s:
D
ab
E
11
A B
aa
C
r
s
©Silberschatz, Korth and Sudarshan3.43Database System Concepts
Division Operation (Cont.)Division Operation (Cont.)
Property Let q – r s Then q is the largest relation satisfying q x s r
Definition in terms of the basic algebra operationLet r(R) and s(S) be relations, and let S R
r s = R-S (r) –R-S ( (R-S (r) x s) – R-S,S(r))
To see why R-S,S(r) simply reorders attributes of r
R-S(R-S (r) x s) – R-S,S(r)) gives those tuples t in
R-S (r) such that for some tuple u s, tu r.
©Silberschatz, Korth and Sudarshan3.44Database System Concepts
Assignment OperationAssignment Operation
The assignment operation () provides a convenient way to express complex queries, write query as a sequential program consisting of a series of assignments followed by an expression whose value is displayed as a result of the query.
Assignment must always be made to a temporary relation variable.
Example: Write r s as
temp1 R-S (r)
temp2 R-S ((temp1 x s) – R-S,S (r))
result = temp1 – temp2
The result to the right of the is assigned to the relation variable on
the left of the .
May use variable in subsequent expressions.
©Silberschatz, Korth and Sudarshan3.45Database System Concepts
Example QueriesExample Queries
Find all customers who have an account from at least the “Downtown” and the Uptown” branches. Query 1
CN(BN=“Downtown”(depositor account))
CN(BN=“Uptown”(depositor account))
where CN denotes customer-name and BN denotes
branch-name.
Query 2
customer-name, branch-name (depositor account)
temp(branch-name) ({(“Downtown”), (“Uptown”)})
©Silberschatz, Korth and Sudarshan3.46Database System Concepts
Find all customers who have an account at all branches located in Brooklyn city.
customer-name, branch-name (depositor account)
branch-name (branch-city = “Brooklyn” (branch))
Example QueriesExample Queries
©Silberschatz, Korth and Sudarshan3.47Database System Concepts
Extended Relational-Algebra-OperationsExtended Relational-Algebra-Operations
Generalized Projection
Outer Join
Aggregate Functions
©Silberschatz, Korth and Sudarshan3.48Database System Concepts
Generalized ProjectionGeneralized Projection
Extends the projection operation by allowing arithmetic functions to be used in the projection list.
F1, F2, …, Fn(E)
E is any relational-algebra expression
Each of F1, F2, …, Fn are are arithmetic expressions involving constants and attributes in the schema of E.
Given relation credit-info(customer-name, limit, credit-balance), find how much more each person can spend:
customer-name, limit – credit-balance (credit-info)
©Silberschatz, Korth and Sudarshan3.49Database System Concepts
Aggregate Functions and OperationsAggregate Functions and Operations
Aggregation function takes a collection of values and returns a single value as a result.
avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values
Aggregate operation in relational algebra
G1, G2, …, Gn g F1( A1), F2( A2),…, Fn( An) (E)
E is any relational-algebra expression
G1, G2 …, Gn is a list of attributes on which to group (can be empty)
Each Fi is an aggregate function
Each Ai is an attribute name
©Silberschatz, Korth and Sudarshan3.50Database System Concepts
Aggregate Operation – ExampleAggregate Operation – Example
Relation r:
A B
C
773
10
g sum(c) (r)sum-C
27
©Silberschatz, Korth and Sudarshan3.51Database System Concepts
Aggregate Operation – ExampleAggregate Operation – Example
Relation account grouped by branch-name:
branch-name g sum(balance) (account)
branch-name account-number balance
PerryridgePerryridgeBrightonBrightonRedwood
A-102A-201A-217A-215A-222
400900750750700
branch-name balance
PerryridgeBrightonRedwood
13001500700
©Silberschatz, Korth and Sudarshan3.52Database System Concepts
Aggregate Functions (Cont.)Aggregate Functions (Cont.)
Result of aggregation does not have a name Can use rename operation to give it a name
For convenience, we permit renaming as part of aggregate operation
branch-name g sum(balance) as sum-balance (account)
©Silberschatz, Korth and Sudarshan3.53Database System Concepts
Outer JoinOuter Join
An extension of the join operation that avoids loss of information.
Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join.
Uses null values: null signifies that the value is unknown or does not exist
All comparisons involving null are (roughly speaking) false by definition.
Will study precise meaning of comparisons with nulls later
©Silberschatz, Korth and Sudarshan3.54Database System Concepts
Outer Join – ExampleOuter Join – Example
Relation loan
loan-number amount
L-170L-230L-260
300040001700
Relation borrower
customer-name loan-number
JonesSmithHayes
L-170L-230L-155
branch-name
DowntownRedwoodPerryridge
©Silberschatz, Korth and Sudarshan3.55Database System Concepts
Outer Join – ExampleOuter Join – Example
Inner Join
loan Borrower
loan borrower
Left Outer Join
loan-number amount
L-170L-230
30004000
customer-name
JonesSmith
branch-name
DowntownRedwood
loan-number amount
L-170L-230L-260
300040001700
customer-name
JonesSmithnull
branch-name
DowntownRedwoodPerryridge
©Silberschatz, Korth and Sudarshan3.56Database System Concepts
Outer Join – ExampleOuter Join – Example Right Outer Join loan borrower
loan-number amount
L-170L-230L-155
30004000null
customer-name
JonesSmithHayes
loan-number amount
L-170L-230L-260L-155
300040001700null
customer-name
JonesSmithnullHayes
loan borrower
Full Outer Join
branch-name
DowntownRedwoodnull
branch-name
DowntownRedwoodPerryridgenull
©Silberschatz, Korth and Sudarshan3.57Database System Concepts
Null ValuesNull Values
It is possible for tuples to have a null value, denoted by null, for some of their attributes
null signifies an unknown value or that a value does not exist.
The result of any arithmetic expression involving null is null.
Aggregate functions simply ignore null values Is an arbitrary decision. Could have returned null as result instead.
We follow the semantics of SQL in its handling of null values
For duplicate elimination and grouping, null is treated like any other value, and two nulls are assumed to be the same Alternative: assume each null is different from each other
Both are arbitrary decisions, so we simply follow SQL
©Silberschatz, Korth and Sudarshan3.58Database System Concepts
©Silberschatz, Korth and Sudarshan3.59Database System Concepts
Null ValuesNull Values
Comparisons with null values return the special truth value unknown If false was used instead of unknown, then not (A < 5)
would not be equivalent to A >= 5
Three-valued logic using the truth value unknown: OR: (unknown or true) = true,
(unknown or false) = unknown (unknown or unknown) = unknown
AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown
NOT: (not unknown) = unknown In SQL “P is unknown” evaluates to true if predicate P evaluates
to unknown
Result of select predicate is treated as false if it evaluates to unknown
©Silberschatz, Korth and Sudarshan3.60Database System Concepts
Modification of the DatabaseModification of the Database
The content of the database may be modified using the following operations: Deletion
Insertion
Updating
All these operations are expressed using the assignment operator.
©Silberschatz, Korth and Sudarshan3.61Database System Concepts
DeletionDeletion
A delete request is expressed similarly to a query, except instead of displaying tuples to the user, the selected tuples are removed from the database.
Can delete only whole tuples; cannot delete values on only particular attributes
A deletion is expressed in relational algebra by:
r r – E
where r is a relation and E is a relational algebra query.
©Silberschatz, Korth and Sudarshan3.62Database System Concepts
Deletion ExamplesDeletion Examples
Delete all account records in the Perryridge branch.
account account – branch-name = “Perryridge” (account)
Delete all loan records with amount in the range of 0 to 50
loan loan – amount 0and amount 50 (loan)
Delete all accounts at branches located in Needham.
r1 branch-city = “Needham” (account branch)
r2 branch-name, account-number, balance (r1)
r3 customer-name, account-number (r2 depositor)
account account – r2
depositor depositor – r3
©Silberschatz, Korth and Sudarshan3.63Database System Concepts
InsertionInsertion
To insert data into a relation, we either: specify a tuple to be inserted
write a query whose result is a set of tuples to be inserted
in relational algebra, an insertion is expressed by:
r r E
where r is a relation and E is a relational algebra expression.
The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.
©Silberschatz, Korth and Sudarshan3.64Database System Concepts
Insertion ExamplesInsertion Examples
Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch.
account account {(“Perryridge”, A-973, 1200)}
depositor depositor {(“Smith”, A-973)}
Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.
r1 (branch-name = “Perryridge” (borrower loan))
account account branch-name, account-number,200 (r1)
depositor depositor customer-name, loan-number,(r1)
©Silberschatz, Korth and Sudarshan3.65Database System Concepts
UpdatingUpdating
A mechanism to change a value in a tuple without charging all values in the tuple
Use the generalized projection operator to do this task
r F1, F2, …, FI, (r)
Each F, is either the ith attribute of r, if the ith attribute is not updated, or, if the attribute is to be updated
Fi is an expression, involving only constants and the attributes of r, which gives the new value for the attribute
©Silberschatz, Korth and Sudarshan3.66Database System Concepts
Update ExamplesUpdate Examples
Make interest payments by increasing all balances by 5 percent.
account AN, BN, BAL * 1.05 (account)
where AN, BN and BAL stand for account-number, branch-name and balance, respectively.
Pay all accounts with balances over $10,0006 percent interest and pay all others 5 percent
account AN, BN, BAL * 1.06 ( BAL 10000 (account))
AN, BN, BAL * 1.05 (BAL 10000 (account))
©Silberschatz, Korth and Sudarshan3.67Database System Concepts
©Silberschatz, Korth and Sudarshan3.68Database System Concepts
R(A B C D)
1 2 2 1 Functional Dependency Graph
2 1 2 3 A B C D
1 2 4 2
3 1 2 1
3 1 1 4
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